Question: $ E = \left[\begin{array}{rr}-1 & 0 \\ 2 & -2\end{array}\right]$ $ A = \left[\begin{array}{rr}1 & 3 \\ 0 & 2\end{array}\right]$ What is $ E A$ ?
Answer: Because $ E$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E A = \left[\begin{array}{rr}{-1} & {0} \\ {2} & {-2}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{3} \\ {0} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{0}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{0}\cdot{0} & ? \\ {2}\cdot{1}+{-2}\cdot{0} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{1}+{0}\cdot{0} & {-1}\cdot\color{#DF0030}{3}+{0}\cdot\color{#DF0030}{2} \\ {2}\cdot{1}+{-2}\cdot{0} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{1}+{0}\cdot{0} & {-1}\cdot\color{#DF0030}{3}+{0}\cdot\color{#DF0030}{2} \\ {2}\cdot{1}+{-2}\cdot{0} & {2}\cdot\color{#DF0030}{3}+{-2}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-1 & -3 \\ 2 & 2\end{array}\right] $